We'll be road-tripping it to get there, which should be some mixture of fun and wow-this-is-really-far-what-were-we-thinking. We've actually driven together almost as far a couple times before (just to practice [that was a joke {if you have to say it was a joke, was it still a joke?}]). We'll pass the time by reading to each other or listening to audio books (so if you know any good ones, feel free to suggest them).

We actually could have stayed in the dorms, which would have been a little funny. Sort of like going back to college, except I never had any girls in my room when I was in college. But we weren't too excited about co-ed bathrooms (or the rest of the dorm experience, really), so we managed to find a deal on priceline to stay at a Fairfield for barely more than the dorms.

I asked Sharayah what other details of the trip I should include, and she came over, hi-jacked my keyboard, and added this bit:

*Another rather important detail is that since it's going to be such a long time away in the middle of summer (the prime tree growing part of the year), we're going to be bringing Mario and Luigi along with us in the back [on the top?] of the car [for sunlighting and photosynthesizing and such]. I don't really think it's a good idea, but Sharayah is the wisest of them all and she thinks it's a Go. So. To Wyoming they go!*

Well, as fun as that would be (and I do regret that we'll miss two weeks of prime summer tree-growth), I wouldn't count on it. As funny as it would be to bring our pet trees on a road trip, I'm not sure they would get enough (any) sunlight in a hotel room for 2 weeks. A missed tree is better than a sad, sunless tree.

Speaking of maths, it's turned out to be quite difficult to prove Brouwer's conjecture on the Laplacian eigenvalues of graphs for

*k*= 3 (that is, that the sum of the largest 3 eigenvalues of the Laplacian matrix of any graph is at most the number of edges of the graph plus 6). I've been proving it for case after case of graphs for weeks, but it seems like it's never-ending. There doesn't seem to be a nice proof that covers many cases at the same time, so I have to keep breaking it down into more and more sub-cases until it can finally be done. That's no problem except the more you break it down, the more cases you have to take care of and the longer it takes. Still, with only two exceptions, I haven't come up to a case for which I couldn't prove the conjecture eventually. For the record, the problem cases are when you take a subgraph of the complete bipartite graph with 3 vertices in one partition and the rest of the vertices in the other, then add either 1 or 2 edges among the vertices in the partition of 3 (if you add no edges or all 3 edges, I can prove it). So, if you could solve that for me, that'd be great. Other than those 2, I've proved it for any graphs with matching numbers 1, 2, or 3, and many of the cases with matching number 4. I only have to consider graphs with matching number at most 5, since a graph with containing 6 independent edges has a subgraph (namely, those 6 edges) whose largest 3 eigenvalues sum to 6, which is the number of edges in that subgraph, and we can show that if a counterexample to Brouwer's conjecture exists, then one exists such that that doesn't happen for any of its subgraphs.

</gibberish>

Ok, sorry, had to get that out of my system. I think that's enough for now.

*be my escape*

## No comments:

## Post a Comment